Chapter 13.3, Problem 13.151P

To determine

(a)

The velocity of the plate immediately after the impact.

Answer to Problem 13.151P

The velocity of the plate immediately after the impact is $\underset{\_}{\underset{\_}{v_p=1.69 m{s}^{-1}}}$.

Explanation of Solution

Given information:

Speed of the ball $=v_x=\overrightarrow{2 }m{s}^{-1}$

Mass of the ball $=m=75g$

Mass of the plate $=M=400g$

General impulse-momentum principal,

$m{v}_1+\sum Im{p}_{1\to 2}=m{v}_2$

Newton’s equation of motion,

$v^2=u^2+2aS$

Calculation:

The velocity of the ball just before impact can be calculated as follows:

Newton’s equation of motion,

$\downarrow v^2=u^2+2aS$

$\downarrow v_y{}^2=0^2+(2\times 9.81 m s^{-2}\times 1.6 m)$

$\downarrow v_y=5.6 m s^{-1}$

$\to v_x=2 m s^{-1}$

The velocity of the ball just after impact can be calculated as follows:

Newton’s equation of motion,

$\downarrow v^2=u^2+2aS$

$\uparrow 0^2=v_{y_1}{}^2-(2\times 9.81 m s^{-2}\times 0.6 m)$

$\uparrow v_{y_1}=3.43 m s^{-1}$

$\to v_{x_1}=2 m s^{-1}$

Applying general impulse-momentum principal,

$\downarrow m{v}_1+\sum Im{p}_{1\to 2}=m{v}_2$

$\downarrow m{v}_y+0=-m{v}_{y_1}+M{v}_p$

$\downarrow v_p=\frac{m(v_y+v_{y_1})}{M}$

$\downarrow v_p=\frac{75\times {10}^{-3}kg(5.6 m s^{-1}+3.43 m s^{-1})}{400\times {10}^{-3}kg}$

$\downarrow v_p=1.69 m s^{-1}$

Conclusion:

The velocity of the plate immediately after the impact is $\underset{\_}{\underset{\_}{v_p=1.69 m{s}^{-1}}}$.

To determine

(b)

The energy loss due to the impact.

Answer to Problem 13.151P

The energy loss due to the impact is $\underset{\_}{\underset{\_}{\Delta E=0.17J}}.$

Explanation of Solution

Given information:

Speed of the ball $=v_x=\overrightarrow{2 }m{s}^{-1}$

Mass of the ball $=m=75g$

Mass of the plate $=M=400g$

General impulse-momentum principal,

$m{v}_1+\sum Im{p}_{1\to 2}=m{v}_2$

Newton’s equation of motion,

$v^2=u^2+2aS$

Kinetic energy,

$K.E=\frac{1}{2}m{v}^2$

Calculation:

Velocity just before impact,

$\downarrow v_y=5.6 m s^{-1}$

$\to v_x=2 m s^{-1}$

$\searrow v_{B_1}=\sqrt{v_x{}^2+v_y{}^2}$

$\searrow v_{B_1}=\sqrt{{(2 m s^{-1})}^2+{(5.6 m s^{-1})}^2}=5.95 m s^{-1}$

Kinetic energy just before impact,

${(K.E)}_1=\frac{1}{2}m{v}_{B_1}{}^2$

${(K.E)}_1=\frac{1}{2}\times 75\times {10}^{-3}kg\times {(5.95 m s^{-1})}^2$

${(K.E)}_1=1.33 J$

Velocity just after impact,

$\uparrow v_{y_1}=3.43 m s^{-1}$

$\to v_{x_1}=2 m s^{-1}$

$\nearrow v_{B_2}=\sqrt{v_{x_1}{}^2+v_{y_1}{}^2}$

$\searrow v_{B_1}=\sqrt{{(2 m s^{-1})}^2+{(3.43 m s^{-1})}^2}=3.97 m s^{-1}$

Kinetic energy just before impact,

${(K.E)}_2=\frac{1}{2}m{v}_{B_2}{}^2+\frac{1}{2}M{v}_p{}^2$

${(K.E)}_2=\frac{1}{2}\times \left[75\times {10}^{-3}kg\times {(5.95 m s^{-1})}^2+400\times {10}^{-3}kg\times {(1.69 m s^{-1})}^2\right]$

${(K.E)}_2=1.16 J$

Conclusion:

The energy loss due to the impact,

$\begin{array}{l}\Delta E={(}_K-{(}_{K}1.16 J\\ \Delta E=1.33 J-1.16 J\end{array}$

$\underset{\_}{\underset{\_}{\Delta E=0.17J}}$